A projectile is fired from a cliff 220 ft above water at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 65 ft per secound. the height ,h, of the projectile abore water is given, h(x)=(-32x^2)/ ((65)^2 ) +x+220. x is the horizontal distance of the projectile from the face of the cliff. What is the maximum value of x?

Answer:248.79 ftStep-by-step explanation:A projectile is fired from a cliff 220 ft above water at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 65 ft per second.[tex]h(x)=-\dfrac{32x^2}{65^2}+x+220[/tex]For maximum value of x, h(x)≥0[tex]-\dfrac{32x^2}{65^2}+x+220\geq0[/tex]Solve quadratic equation for x [tex]-\dfrac{1}{4225}(32x^2-4225x-929500)\geq0[/tex][tex]32x^2-4225x-929500\leq0[/tex]Using quadratic formula,[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]where, a=32, b=-4225, c=-929500[tex]x=\dfrac{4225\pm\sqrt{4225^2-4(32)(-929500)}}{2(32)}[/tex][tex]x\geq-116.75\text{ and }x\leq 248.79[/tex]Hence, The maximum value of x will be 248.79 ft